\(n_{CO_2}=\dfrac{PV}{RT}=\dfrac{4.928\cdot2}{0.082\cdot\left(273+27.3\right)}=0.4\left(mol\right)\)
\(\)\(CaCO_3\cdot MgCO_3\underrightarrow{^{t^0}}MgO+CaO+CO_2\)
\(0.4.................................................0.4\)
\(m=0.4\cdot\left(100+84\right)=73.6\left(g\right)\)
\(\%CaCO_3\cdot MgCO_3=\dfrac{73.6}{92}\cdot100\%=80\%\)